Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename A>
ostream &operator<<(ostream &cout, vector<A> const &v) {
  for (long long i = 0; i < v.size(); i++) cout << v[i] << " ";
  cout << "\n";
}
template <typename A, typename B>
ostream &operator<<(ostream &cout, map<A, B> const &v) {
  for (auto x : v) cout << x.first << ": " << x.second << "\n";
}
template <typename A>
ostream &operator<<(ostream &cout, vector<array<A, 2>> const &v) {
  for (long long i = 0; i < v.size(); i++)
    cout << v[i][0] << " " << v[i][1] << "\n";
}
int32_t main() {
  long long m, n;
  cin >> m >> n;
  vector<long long> row(m, 501), col(n, 501);
  long long a[100][100], b[100][100];
  for (long long i = 0; i < m; i++) {
    for (long long j = 0; j < n; j++) cin >> a[i][j], b[i][j] = a[i][j];
  }
  for (long long i = 0; i < m; i++) {
    long long mn = 1e9;
    for (long long j = 0; j < n; j++) mn = min(mn, a[i][j]);
    row[i] = mn;
  }
  vector<string> ans1, ans2;
  for (long long i = 0; i < m; i++) {
    for (long long j = 0; j < row[i]; j++)
      ans1.push_back("row " + to_string(i + 1));
    for (long long j = 0; j < n; j++) a[i][j] -= row[i];
  }
  for (long long j = 0; j < n; j++) {
    long long mn = 1e9;
    for (long long i = 0; i < m; i++) mn = min(mn, a[i][j]);
    col[j] = mn;
  }
  for (long long j = 0; j < n; j++) {
    for (long long i = 0; i < col[j]; i++)
      ans1.push_back("col " + to_string(j + 1));
    for (long long i = 0; i < m; i++) a[i][j] -= col[j];
  }
  for (long long i = 0; i < m; i++) {
    for (long long j = 0; j < n; j++) {
      if (a[i][j]) {
        cout << -1;
        return 0;
      }
    }
  }
  for (long long j = 0; j < n; j++) {
    long long mn = 1e9;
    for (long long i = 0; i < m; i++) mn = min(mn, b[i][j]);
    col[j] = mn;
  }
  for (long long j = 0; j < n; j++) {
    for (long long i = 0; i < col[j]; i++)
      ans2.push_back("col " + to_string(j + 1));
    for (long long i = 0; i < m; i++) b[i][j] -= col[j];
  }
  for (long long i = 0; i < m; i++) {
    long long mn = 1e9;
    for (long long j = 0; j < n; j++) mn = min(mn, b[i][j]);
    row[i] = mn;
  }
  for (long long i = 0; i < m; i++) {
    for (long long j = 0; j < row[i]; j++)
      ans2.push_back("row " + to_string(i + 1));
    for (long long j = 0; j < n; j++) b[i][j] -= row[i];
  }
  if (int32_t(ans1.size()) <= int32_t(ans2.size())) {
    cout << ans1.size() << "\n";
    for (auto x : ans1) cout << x << "\n";
  } else {
    cout << ans2.size() << "\n";
    for (auto x : ans2) cout << x << "\n";
  }
}